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  • Leetcode: Minimum Size Subarray Sum

    Given an array of n positive integers and a positive integer s, find the minimal length of a subarray of which the sum ≥ s. If there isn't one, return 0 instead.
    
    For example, given the array [2,3,1,2,4,3] and s = 7,
    the subarray [4,3] has the minimal length under the problem constraint.
    
    click to show more practice.
    
    More practice:
    If you have figured out the O(n) solution, try coding another solution of which the time complexity is O(n log n).

    Haven't think about the O(nlogn) solution.

    O(n) solution is to maintain a window

     1 public class Solution {
     2     public int minSubArrayLen(int s, int[] nums) {
     3         int minWin = Integer.MAX_VALUE;
     4         int l=0, r=0;
     5         if (nums==null || nums.length==0) return 0;
     6         int sum = 0;
     7         while (r < nums.length) {
     8             sum += nums[r];
     9             while (sum >= s) {
    10                 minWin = Math.min(minWin, r-l+1);
    11                 sum -= nums[l++];
    12             }
    13             r++;
    14         }
    15         return minWin == Integer.MAX_VALUE? 0 : minWin;
    17     }
    18 }

    My solution: shrink as much as possible while maintaining the window always correct

     1 public class Solution {
     2     public int minSubArrayLen(int s, int[] nums) {
     3         int minWin = Integer.MAX_VALUE;
     4         int l=0, r=0;
     5         if (nums == null || nums.length == 0) return 0;
     6         int sum = 0;
     7         while (r < nums.length) {
     8             sum += nums[r];
     9             while (sum - nums[l] >= s) {
    10                 sum -= nums[l++];
    11             }
    12             if (sum >= s) 
    13                 minWin = Math.min(minWin, r - l + 1);
    14             r++;
    15         }
    16         
    17         return minWin == Integer.MAX_VALUE? 0 : minWin;
    18     }
    19 }
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  • 原文地址:https://www.cnblogs.com/EdwardLiu/p/5052889.html
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